Saturday, December 26, 2020

Harmonic mean of integers

The harmonic mean of a set of $n$ numbers $x_i$ is defined as $\frac{n}{\sum_{i=1}^n x_i^{-1}}$. While investigating the number of subsets of $\{1,...,n\}$ such that the harmonic mean is an integer (OEIS sequence A339453), I formulated and proved the following result, which states that for $n >1$ positive integers whose maximum is a prime power that is attained by a single element, their harmonic mean is not an integer:

Theorem: Let $x_i$ be a finite set of positive integers such that $x_j = \max_i x_i = p^k$ for some prime $p$ and positive integer $k$ and all other numbers $x_i$ are strictly less than $p^k$, then the harmonic mean of $\{x_i\}$ is not an integer.

Proof: Let $M$ be the least common multiple of $\{x_i\}$. Assume that $x_i$ are sorted in nondecreasing order. Thus $x_n = p^k$ and $x_i < p^k$ for all $i<n$ . Then $M = Wp^k$ where $p$ does not divide $W$. Let $Q_i = M/x_i$ and $Q = \sum_i Q_i$. This implies that $Q_n = W$ and $p$ divides $Q_i$ for $i <n$.

The harmonic mean $H$ can then be written as $nM/Q$. Since $p$ does not divide $W$, this implies that $p$ does not divide $Q$. Suppose $H$ is an integer. Then this implies that $Q$ divides $nM/p^k = nW$.

As $x_i < x_n$ for $i < n$, this implies that $Q_i > W$ for $i < n$, i.e. $Q > nW$, and this contradicts the fact that $Q$ divides $nW$ and thus $H$ is not an integer.

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